2001 AIME I Problems/Problem 7
Problem
Triangle has , and . Points and are located on and , respectively, such that is parallel to and contains the center of the inscribed circle of triangle . Then , where and are relatively prime positive integers. Find .
Contents
Solution 1
Let be the incenter of , so that and are angle bisectors of and respectively. Then, so is isosceles, and similarly is isosceles. It follows that , so the perimeter of is . Hence, the ratio of the perimeters of and is , which is the scale factor between the two similar triangles, and thus . Thus, .
Solution 2
The semiperimeter of is . By Heron's formula, the area of the whole triangle is . Using the formula , we find that the inradius is . Since , the ratio of the heights of triangles and is equal to the ratio between sides and . From , we find . Thus, we have
Solving for gives so the answer is .
Or we have the area of the triangle as . Using the ratio of heights to ratio of bases of and from that it is easy to deduce that .
Solution 3 (mass points)
Let be the incenter; then it is be the intersection of all three angle bisectors. Draw the bisector to where it intersects , and name the intersection .
Using the angle bisector theorem, we know the ratio is , thus we shall assign a weight of to point and a weight of to point , giving a weight of . In the same manner, using another bisector, we find that has a weight of . So, now we know has a weight of , and the ratio of is . Therefore, the smaller similar triangle is the height of the original triangle . So, is the size of . Multiplying this ratio by the length of , we find is . Therefore, .
Solution 4 (Faster)
More directly than Solution 2, we have
Solution 5
Diagram borrowed from Solution 3.
Let the angle bisector of intersects at .
Applying the Angle Bisector Theorem on we have Since is the angle bisector of , we can once again apply the Angle Bisector Theorem on which gives Since we have Solving gets . Thus .
~ Nafer
Solution 6
Let be the foot of the altitude from to and be the foot of the altitude from to . Evidently, where is the inradius, , and is the semiperimeter. So, Therefore, by similar triangles, we have .
Solution 7
Label the point the angle bisector of intersects . First we find and . By the Angle Bisector Theorem, and solving for each using the fact that , we see that and .
Because is the angle bisector of , we can simply calculate it using Stewarts,
Now we can calculate what is. Using the formula to find the distance from a vertex to the incenter, .
Now because , we can find by . Dividing and simplifying, we see that . So the answer is
~YBSuburbanTea
Solution 8 (vectors)
To solve this problem, we can use the fact that, in , the vector representation of the incenter is and that that the vector of the foot of the bisector of on is , where and .
Let point be the origin of the coordinate plane. Then, is the zero vector, so we can simplify our expression for to . Now, note that the vector components of and are the same, but they are multiplied by different scalars. Thus, the ratio of these scalars is the ratio of these vectors' magnitudes. Thus, we have .
Let and . Because , we have . Further, because , we have . Thus, by transitivity, . We know that , so .
Thus, our answer is .
See also
2001 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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